# What is the cardinality of the group of bijections from $\Omega$ to $\Omega$ with finite support?

These questions cropped up in the discussion in this question,

What is the cardinality of the group of bijections from $\Omega$ to $\Omega$ with finite support, where $\Omega=\mathbb{N}$?

What about $\Omega$ a set with cardinality greater than that of the natural numbers? That is, is the cardinality of $S_{\Omega}$ equal to, or greater than, that of $\Omega$?

If one takes the set of bijections without finite support then you get that this group has cardinality strictly greater than that of $\Omega$. However, I am unsure what to do for the set of bijections with finite support (although I have posted an idea in the discussion in the linked-to question for $\Omega=\mathbb{N}$).

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For any infinite cardinal $\kappa$, it is $\kappa$. – André Nicolas Nov 1 '11 at 17:39
@AndréNicolas: Yes...but why?!? – user1729 Nov 1 '11 at 17:41
Any bijection with finite support lies in some $S_n\subset S_{\mathbb{N}}$, and conversely. So the subset of bijections with finite support is a numerable union of finite sets, thus numerable. More generally, the same holds for infinite $\Omega$ with a union taken over all finite subsets of $\Omega$, which has the same cardinality as $\Omega$ (I think some form of choice is needed to say that $\Omega\simeq \Omega\times\cdots\times \Omega$ with a finite amount of copies of $\Omega$ though I could be wrong), so there are $|\Omega|$ bijections with finite support. – Olivier Bégassat Nov 1 '11 at 17:44
@user1729: The answer has been given by Olivier Bégassat. If $\Omega$ has cardinality $\kappa$, and $n$ is an integer, the set of $n$-subsets of $\Omega$ has cardinality $\kappa$. We do need the Axiom of Choice. – André Nicolas Nov 1 '11 at 18:23
@OlivierBégassat: The assertion that $\Omega\times\Omega$ can be bijected with $\Omega$ for any infinite set $\Omega$ is equivalent to the Axiom of Choice (by a theorem of Tarski). – Arturo Magidin Nov 1 '11 at 19:38

I will assume the Axiom of Choice throughout.

Let $X$ be any infinite set, with $|X|=\kappa$, and for any $n\in\mathbb{N}$, let $X^n$ be the set of ordered $n$-tuples.

There is a surjection from $X^n$ to the collection of all subsets of $X$ with at most $n$ elements, namely the map that takes the $n$-tuple $(x_1,\ldots,x_n)$ to the subset $\{x_1,\ldots,x_n\}$. That means that $$\mathcal{P}_n(X) = \{A\subseteq X\mid |A|=n\}$$ has cardinality at most that of $X^n$.

But for infinite cardinals $\kappa$, $\kappa\times\kappa = \kappa$ if we assume the Axiom of Choice (in fact, this is equivalent to AC), from which it follows by induction that $\kappa^n = \kappa$ for all $n\in\mathbb{N}$. So $|\mathcal{P}_n(X)| \leq |X^n| = \kappa$. Moreover, $\kappa\leq |\mathcal{P}_n(X)|$ (biject $X$ with $X-\{x_1,\ldots,x_{n-1}\}$ for some subset of exactly $n-1$ elements to get that there are at least $|X|$ subsets with $n$ elements). So $|\mathcal{P}_n(X)| = |X|=\kappa$.

Thus, if we let $$\mathcal{P}_{\lt\infty}(X) = \{A\subseteq X \mid |A|\lt\aleph_0\},$$ then \begin{align*} \left|\mathcal{P}_{\lt\infty}(X)\right| &= \left|\bigcup_{n\in\mathbb{N}}\mathcal{P}_n(X)\right|\\ &= \sum_{n\in\mathbb{N}}\left|\mathcal{P}_n(X)\right|\\ &\leq \sum_{n\in \mathbb{N}}\kappa\\ &= |\mathbb{N}|\kappa = \aleph_0\kappa = \kappa, \end{align*} with the last equality because if we assume the Axiom of Choice, and $\kappa$ and $\lambda$ are any two cardinals, at least one infinite and neither one equal to $0$, then $\kappa+\lambda = \kappa\lambda = \max\{\kappa,\lambda\}$.

Now, for each $A\in \mathcal{P}_{\lt\infty}(X)$, the subgroup of $S_X$ with support contained in $A$ has $n!$ elements, where $|A|=n$. Therefore, \begin{align*} |S_X| &\leq \sum_{A\in\mathcal{P}_{\lt\infty}(X)} |S_A|\\ &\leq \sum_{A\in\mathcal{P}_{\lt\infty}(X)} \aleph_0\\ &= |P_{\lt\infty}(X)|\aleph_0 = \kappa\aleph_0 = \kappa. \end{align*} On the other hand, if we fix $x\in X$ and consider all transpositions of the form $(x,y)$ with $y\in X-\{x\}$, then we have that $|S_X$ has at least $|X|=\kappa$ elements. Therefore, $|S_X| = \kappa$.

That is: if $X$ is infinite, then the group of permutations of $X$ with finite support has cardinality $|X|$.

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This is hardly worth pointing out, but the same argument goes through without choice under the additional hypothesis that $X$ is wellorderable. – user83827 Nov 1 '11 at 20:56
@Arturo: I gave proof for the fact that $\kappa\times\kappa=\kappa$ for well ordered cardinals on this answer, which is somewhat dual to that comment on the original question. (This is the argument which ccc refers to) – Asaf Karagila Nov 1 '11 at 21:31

There are countably many finite subsets of $\mathbb{N}$. Moreover if we consider ordered finite subsets we still get a countable number of them. We construct all bijections with finite support by choosing an ordered finite subset $A \subset \mathbb{N}$ and an ordered finite subset $B \subset \mathbb{N}$ and letting $\phi$ map $a_k$ to $b_k$ and be identity elsewhere. Thus there are countably many such bijections.

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The case of $\mathbb N$ requires no axiom of choice, and we can prove it to be countable. Here is how:

Note that $\pi\colon\mathbb N\to\mathbb N$ is a permutation with a finite support if and only if there exists some $k$ such that $\pi(n)=n$ for all $n>k$. Therefore for every such $\pi$ there is some $k(\pi)$ such that $\pi$ is actually a permutation of $\{0,\ldots,k-1\}$.

Now since every for every $k$ there are only finitely many permutations of $\{0,\ldots,k-1\}$ we would like to conclude that this is a countable union of finite sets and therefore countable. Assuming the axiom of choice (or even axiom of choice for finite sets) would conclude this portion of the proof.

Without the axiom of choice, however, we still have some work a head of us.

Let us define the following sequence of sets

$$A_k = \{\pi\mid k(\pi)=k\}$$ That is all the permutations that $k$ is the maximal $k$ moved.

Note that for $m<k$ we have $A_m\subsetneq A_k$. Now we will define inductively an enumeration of $A=\bigcup_{k\in\mathbb N} A_k$.

1. For $k=2$ we have the first nonempty $A_k$, and it only has one element so we are done.

2. Suppose $A_k$ was ordered by $<_k$, we define the order $<_{k+1}$ on $A_{k+1}$ as follows, for any $\tau,\pi\in A_k$:

$$\pi<_{k+1}\tau\iff \begin{cases} &\pi,\tau\in A_k&\land&\pi<_k\tau &\text{ or}\\ &\pi\in A_k&\land&\tau\notin A_k &\text{ or}\\ &\pi,\tau\notin A_k&\land&\pi(n)<\tau(n)\text{ where } n=\min\{t<k\mid \pi(t)\neq\tau(t)\} \end{cases}$$

Let us verify that this is indeed a linear order, for any $\pi,\tau\in A_{k+1}$ either both in $A_k$, or at one of them is not in which cases it is clear that the two permutations are comparable, and if both of them in $A_{k+1}$ then there is some $n$ for which $\pi(n)\neq\tau(n)$ or else $\pi=\tau$.

It is also clear why $<_{k+1}$ when restricted to $A_k$ is exactly $<_k$. Lastly since every finite linear order is a well-order this gives us an increasing chain of linear orders on increasing subsets of $A$.

Let $\prec=\bigcup_k <_k$, then $\prec$ is a well ordering of $A$, and it is countable as it is the limit of finite well-orderings. Therefore $A$ has a countable enumeration by $\prec$. $\square$

A note on the general case without the axiom of choice:

While assuming the axiom of choice we may replace $\mathbb N$ at the above proof by any other set, and the proof will only change by a little bit. We could in fact generalize it without much trouble either to include even larger support for larger sets.

However, in the absence of choice things are not that pretty. Without the axiom of choice we have a may have an infinite set $S$ such that $f\colon S\to S$ is bijective then it has a finite support. However the set of all permutations of $S$ will have a strictly larger cardinality than that of $S$.

Assuming, though, that $S$ can be well-ordered then things are fine again, and the above proof works (changing the necessary details for it to fit $|S|$).

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NOTE: I completely misread the original post, fairly egregiously too. I leave this up just in the small case it may be useful. @Moderators, feel free to remove this post if it's inappropriate.

I remember last year I tried to sit down and prove this for myself (it is often stated, but not often proved--for example, it comes up a lot when talking about dimensions of infinite dimensional vector spaces). So, for what it's worth here's my intuitive (to MYSELF) way to do things. It seems fairly close to what Arturo did, at least in idea, not as much in actual proof. The idea is to basically show that the set $\mathscr{P}_0(X)$ (if $X$ is our infinite set) of all finite subsets is just a countable union of $X^n$ (or things equipotent to it) and then to note that each $X^n$ is equipotent to $X$ and so the countable union should be equipotent to $X$. So, here we go

(NB: I am by no means any kind of logician/set theorist, so take this with a grain of foundations-salt. This is no idle threat either, I ran this by a professor of mine, and he thought it seemed ok, but I am still not positive.)

What we claim is that is $\mathscr{S}_n=\left\{A\subseteq X:\#(A)=n\right\}$ that $\#(\mathscr{S}_n)=\#(X)$ for all $n\in\mathbb{N}$. We do this by Schroeder-Bernstein. The one way is obvious since if we let $C_n=\left\{(a_1,\cdots,a_n)\in X^n:a_i\ne a_j\; i\ne j\right\}$ then $f:C_n\to\mathscr{S}_n:(a_1,\cdots,a_n)\mapsto \{a_1,\cdots,a_n\}$ is a surjection and so $\#(\mathscr{S}_n)\leqslant \#(C_n)\leqslant \#(X^n)=\#(X)$ (where we used the obvious facts that $C_n\subseteq X^n$ and $\#(X^n)=\#(X)$ since $X$ is infinite).

To prove the reverse inequality we well-order $X$ with some ordering $\preceq$. We then note that trivially there exists some finite subset $V$ of $X$ such that every element $X-V$ has at least $n-1$ successors with $\preceq$ (this step is unnecessary technically because we could just replace $X$ with a different ordinal, but, I think from a prerequisite point of view, this is simpler). We then define $f:X\to\mathscr{S}_n$ by $f:x\mapsto\{x,x+1,\cdots,x+(n-1)\}$. It's not hard to see that $f$ is an injection and so $\#(X)\leqslant \#(\mathscr{S}_n)$.

Thus, from Schroeder-Bernstein we may conclude that $\#(X)=\#(\mathscr{S}_n)$ and the rest of the proof follows very similar to Arturo's. Namely: $\displaystyle \#(\mathscr{P}_0)=\#\left(\bigsqcup_{n\in\mathbb{N}}\mathscr{S}_n\right)=\sum_{n\in\mathbb{N}}\#(\mathscr{S}_n)=\aleph_0\#(X)=\#(X)$ where we used the fact that $\aleph_0\#(X)=\max\{\#(X),\aleph_0\}$ and $\aleph_0\leqslant\lambda$ for all infinite cardinals $\lambda$.

I hope this helps, and remember my disclaimer!

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